Jekyll2017-02-27T00:37:01+00:00http://mathsalad.com/MathsaladA note on the “trivial group presentation game”: \(I\) has a winning strategy on the group \(F_2\)2015-05-29T00:00:00+00:002015-05-29T00:00:00+00:00http://mathsalad.com/2015/05/29/A-note-on-the-%22tivial-group-game%22<p>In <a href="/2015/05/15/the-strong-measure-zero-game.html">my post on the strong measure zero game</a>, I mentioned a game <script type="math/tex">G_1(\mathcal {N}_H, \mathcal{N}_H)</script>, where player one, who I will call <script type="math/tex">I</script>, presents player two, who I will call <script type="math/tex">II</script>, a subset from a group <script type="math/tex">H</script>, where the normal closure of the set is the full group, and <script type="math/tex">II</script>, chooses one element from each set presented, trying to get a set whose normal closure is again the full group. The game lasts <script type="math/tex">\omega</script> innings (<script type="math/tex">\mathbb{N}= \omega</script> if you prefer that notation). Essentially the idea is that <script type="math/tex">I</script> gives presentations of the trivial group, and <script type="math/tex">II</script> is trying to end up with a presentation of the trivial group.</p>
<p>As a simple example, we can consider the group of integers <script type="math/tex">(\mathbb Z,+)</script>. Since the group is abelian any group that normally generates the group, also generates the group. This also means that there is some linear combination of elements from any presented set that has value <script type="math/tex">1</script>. What player <script type="math/tex">II</script> can do to win, is choose a non identity element <script type="math/tex">x_0</script>, where <script type="math/tex">\vert x_0 \vert</script> is the least among non identity elements. If <script type="math/tex">\vert x_0 \vert=1</script>, player <script type="math/tex">II</script> has already chosen a set that normally generates and may choose arbitrarily. Suppose that <script type="math/tex">II</script> has chosen <script type="math/tex">x_0,...,x_n</script>, if <script type="math/tex">d:=\gcd(x_0,...,x_n)=1</script> we choose an arbitrary element. If we have <script type="math/tex">d>1</script>, then choose an element <script type="math/tex">x_{n+1}</script>, so that <script type="math/tex">% <![CDATA[
\gcd(d,x_{n+1})<d %]]></script>. Note that we can do that, since otherwise, every element in the presented set would have <script type="math/tex">d > 1</script> dividing it, hence the greatest common divisor of the set, without the identity, would be greater than or equal to <script type="math/tex">d</script>, contradicting the set generates the group. So we can always decrease our greatest common divisor till we reach <script type="math/tex">1</script>. So we have a strategy where <script type="math/tex">II</script> will win every time. A similar argument can be used to show that <script type="math/tex">II</script> wins in groups where every strictly ascending chain of infinite index subgroups, is finite.</p>
<p>It is pretty easy to find groups where <script type="math/tex">I</script> wins, by using large groups, like <script type="math/tex">(\mathbb{R},+)</script>, but what about finitely generated groups? We will show, in this post, that <script type="math/tex">I</script> has a winning strategy to win <script type="math/tex">F_2</script>, in fact it will be a “fixed strategy” where <script type="math/tex">I</script> will play the same no matter what <script type="math/tex">II</script> picks on <script type="math/tex">II</script>’s turns.</p>
<p>Our proof will use facts from <a href="https://en.wikipedia.org/wiki/Small_cancellation_theory">small cancellation theory</a>, in particular small cancellation groups are not trivial. I will explain a bit of the idea of small cancellation conditions, and give key definitions. All the small cancellation that we discuss can be found in <em>Combinatorial Group Theory</em> Lyndon and Schupp, or <em>Geometry of Defining Relations in Groups</em> by Ol’shanskii, or <a href="https://en.wikipedia.org/wiki/Small_cancellation_theory">Wikipedia</a>.</p>
<p>The idea of small cancellation theory is that, if you have a set of relations, satisfying a “small cancellation condition”, that is when you multiply the two different words, the length of the reduced product (remove the <script type="math/tex">xx^{-1}</script>) is not much shorter than the the sum of the lengths of the word. This, in a sense, gives control over what words end up representing the identity, and how words can be reduced using the relations that satisfy the small cancellation conditions.</p>
<h2 id="basic-definition-for-small-cancellation-theory">Basic definition for small cancellation theory</h2>
<p>The information in this section is all fairly basic stuff, you may want to skip this section, or just read the <script type="math/tex">C'(1/6)</script> definition and come back if there are some words you do not know.</p>
<p><strong>Basic definitions for words in free groups.</strong> related to words and free groups. A word is a string of elements from some some set called an alphabet. For example <script type="math/tex">x, xyxx^{-1},</script> and <script type="math/tex">xy</script> are all (not equal) words from the alphabet <script type="math/tex">\{ x,y,x^{-1},y^{-1} \}</script>. A word is reduced if there is no occurrence of <script type="math/tex">tt^{-1}</script> or <script type="math/tex">t^{-1}t</script> in the word, and cyclically reduced provided it is not a nontrivial conjugate, so <script type="math/tex">w \neq t w' t^{-1}</script> and <script type="math/tex">w \neq t^{-1} w' t</script>. The length of a word <script type="math/tex">w</script>, which we call <script type="math/tex">\vert w \vert</script> is the number of elements in the reduced version of the word; this is well defined. So <script type="math/tex">\vert xyxx^{-1} \vert = \vert xy \vert =2</script>, for example.</p>
<p><strong>Definition of symmetrized presentation.</strong> Suppose <script type="math/tex">G = \langle X \mid R \rangle</script>. We say the set of relations <script type="math/tex">R</script> is symmetrized provided it is closed under taking inverses, and taking cyclic permutations of words up to being cyclically reduced, and every word in <script type="math/tex">R</script> is cyclical reduced. <script type="math/tex">G</script> has a symmetrized presentation provided <script type="math/tex">R</script> is symmetrized. For example</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
R = &\{ab^3ab^4,b^3ab^4a^{-1}, ab^4 a ^{-1} b^{-3}, \\
& b^4a^{-1} b^{-3} a ^{-1}, a^{-1}b^{-3}a^{-1}b^{-4},... \text{inverses} \}.
\end{align*} %]]></script>
<p><strong>Definition of <script type="math/tex">C'(1/6)</script>:</strong> We say that <script type="math/tex">p</script> is a piece relative to <script type="math/tex">R</script> provided there exists two distinct elements <script type="math/tex">r_1,r_2 \in R</script> with <script type="math/tex">r_1=pc_1</script> and <script type="math/tex">r_2=pc_2</script>, where “<script type="math/tex">=</script>” is word equality, so no cancellation. We way <script type="math/tex">R</script> has condition <script type="math/tex">C'(1/6)</script> if <script type="math/tex">r \in R</script> and <script type="math/tex">r=pc</script> where <script type="math/tex">p</script> is a piece, then <script type="math/tex">% <![CDATA[
\vert p \vert < \frac{1}{6} \vert r \vert %]]></script>. There is an entirely similar definition for <script type="math/tex">C'(\lambda).</script></p>
<p>The reason I only mention <script type="math/tex">C'(1/6)</script> is that it is a strong enough small cancellation condition to guarantee a group is infinite. Note that you can have a group that has a <script type="math/tex">C'(1/6)</script> presentation, but may not be given to you that way. It is an interesting fact that for <script type="math/tex">\lambda > 1/5</script>, every group has a <script type="math/tex">C'(\lambda)</script> presentation. There are many other small cancellation conditions, <script type="math/tex">C(p)</script> and <script type="math/tex">T(p)</script> for example, and you can mix-and-match different conditions too, but that is really beyond this post.</p>
<h2 id="player-i-has-a-winning-strategy-in-g_1-mathcaln_f_2mathcaln_f_2">Player <script type="math/tex">I</script> has a winning strategy in <script type="math/tex">G_1( \mathcal{N}_{F_2},\mathcal{N}_{F_2})</script></h2>
<p>Let <script type="math/tex">F_2</script> be the free group of rank <script type="math/tex">2</script>, on the set <script type="math/tex">\{x,y\}</script>. We will show that <script type="math/tex">F_2</script> has a winning strategy, by showing that there is a sequence of sets, <script type="math/tex">A_0,...,A_n,...</script>, that normally generate, even generates, the group <script type="math/tex">F_2</script>, and any one element choice from each of these sets gives a set <script type="math/tex">\{a_0,...,a_n,...\}</script>, which will correspond to a group, <script type="math/tex">G=\langle x,y \mid a_0,...,a_n,... \rangle</script>, that is <script type="math/tex">C'(1/6)</script>, which means that <script type="math/tex">\{a_0,...,a_n,...\}</script> can not normally generate <script type="math/tex">F_2</script>. The strategy for player <script type="math/tex">I</script> will be to play <script type="math/tex">A_n</script> on the <script type="math/tex">n</script>th turn.</p>
<p>There will be a couple parts where I make some estimates on inequalities, and make no attempt to make these as good as possible(the estimates I make are <em>good enough</em>). If you are sensitive to such disregard of sharp bounds, you will need to have your eyes covered for the rest of this proof. You have been warned.</p>
<p><strong>Proposition 1.</strong> Let</p>
<script type="math/tex; mode=display">w_n=\prod_{i=1}^{9 \cdot 100^{n+1}} xy^{100^{n+1}+i}=xy^{100^{n+1}+1}...xy^{100^{n+2}}.</script>
<p>Then we have that <script type="math/tex">A_n=\{w_{n+1},xw_{n+1},w_{n+1}y\}</script> generates the group <script type="math/tex">F_2</script>, but any one element choice from each <script type="math/tex">A_n</script> results in a set of relations for a <script type="math/tex">C'(1/6)</script> group. In particular this means that for any choice of <script type="math/tex">a_i</script>’s, where <script type="math/tex">a_i \in A_i</script>, that <script type="math/tex">\langle x ,y \mid a_0,...,a_n,... \rangle</script> is an infinite group since it will be <script type="math/tex">C'(1/6)</script>.</p>
<p><em>proof of Lemma 1.</em> First it is clear that each <script type="math/tex">A_n</script> generates <script type="math/tex">F_2</script> since we can multiply <script type="math/tex">w_{n+1}^{-1}</script> to cancel <script type="math/tex">w_{n+1}</script> on each term to end up with <script type="math/tex">x,y</script>, which generates the group. Let <script type="math/tex">R_w</script> be the smallest symmetrized set of words containing the word <script type="math/tex">w</script>, which is all the cyclically reduced, cyclic permutations of the word <script type="math/tex">w</script> and their inverses. For <script type="math/tex">R_{w_n}</script> all pieces are of the form <script type="math/tex">b^i a^{\pm 1} b^j</script>, where <script type="math/tex">-100^{n+2} \leq i,j\leq 100^{n+2}</script> (note that for many <script type="math/tex">i,j</script> can not actually be obtained). So the length of a pieces <script type="math/tex">p</script> are bounded by</p>
<script type="math/tex; mode=display">% <![CDATA[
2 \cdot 100^{n+2}+1<3\cdot 100^{n+2}, %]]></script>
<p>and the length of each <script type="math/tex">w_n</script> is bounded from below by</p>
<script type="math/tex; mode=display">9\cdot 100^{n+1} \cdot 100^{n+1}.</script>
<p>We get that</p>
<script type="math/tex; mode=display">\frac{ \vert p \vert }{ \vert w_n \vert } \le \frac{ 3 \cdot 100^{n+2} }{9 \cdot 100^{2n+2}}= \frac{1}{3 \cdot 100^n}</script>
<p>which for <script type="math/tex">n>0</script> we have the right hand side is less than <script type="math/tex">1/6</script>. A very similar argument can be carried out for <script type="math/tex">R_{w_ny}</script> and <script type="math/tex">R_{xw_n}</script>, and in fact you can use the same bounds, since our bounds are so rough. So each <script type="math/tex">R_w</script>, for <script type="math/tex">w \in A_n</script>, gives a <script type="math/tex">C'(1/6)</script> presentation. Say that <script type="math/tex">m>n</script> and <script type="math/tex">w \in A_m</script>, and <script type="math/tex">u \in A_n</script>, and we consider <script type="math/tex">R_w,R_u</script>. Note, that at “worst” the amount of cancellation between words in <script type="math/tex">R_w</script> and <script type="math/tex">R_u</script> will be bounded by the cancellation bounds we established above for <script type="math/tex">R_u</script>, since elements in <script type="math/tex">R_u</script> have small enough exponents on their <script type="math/tex">y</script>’s that they can not completely cancel out a <script type="math/tex">y^k</script> term in <script type="math/tex">R_w</script>. This worked for arbitrary choices. So let <script type="math/tex">a_0,...,a_n,...</script> be a sequence where <script type="math/tex">a_i \in A_i</script>, then <script type="math/tex">% <![CDATA[
R=\bigcup_{i< \omega} R_{a_i} %]]></script> is a symmetrized set of relations, and we know that between two words, the cancellation is less than <script type="math/tex">1/6</script> the length of either of the words, so it gives a <script type="math/tex">C'(1/6)</script> presentation, hence <script type="math/tex">\langle x,y \mid a_0,...,a_n,... \rangle</script> is a <script type="math/tex">C'(1/6)</script> group, and is infinite. <script type="math/tex">\square</script></p>
<p>I came up with this game in a class I took on selection principles, and studied it as a final project, and at the time I knew very little group theory (I still know very little), and even thought all finitely generated groups had a winning strategy for player <script type="math/tex">II</script>. At the end of the class I only really found some conditions for player <script type="math/tex">II</script> winning in finitely generated groups. It had been at the back of my mind since the class, I would pick it up and work on it for a while before setting down, multiple times. In a way, thinking about this problem and closely related things really influenced my mathematical interests, I have a feeling it will show in future posts. In the end I am actually not to sure these ideas are of interest, at least in the finitely generated group case. I suspect it will be a while before I discuss these games on groups again, but I will end this with a couple questions to ponder:</p>
<ul>
<li>For finitely generated groups is there a nice demarcation line between player <script type="math/tex">I</script> or <script type="math/tex">II</script> having a winning strategy? For groups in general?</li>
<li>Are there groups where there is no winning strategy for either player? Are there finitely generated groups where there is no winning strategy?</li>
<li>There are a couple of other, similar ideas to consider, one where you replace normally generate with simply generate, or replace with generate finite index subgroup, or normally generate finite index subgroup, or mix-and-match the different sets. (I have not really thought about them much, but there are probably a lot of equivalences, because once you reach a finite index subgroup, in many cases, reach the full group) Are there any interesting connections and differences among them?</li>
<li>For those who have looked up the connection of these sorts of games with selection principles and Ramsey statements(see the <a href="/2015/05/15/the-strong-measure-zero-game.html">references in the strong measure zero game post</a>): are there interesting connection between those and the games for finitely generated groups? For countable groups? For groups? In particular what would the Ramsey statements be?</li>
<li>Can you think of any cool group theoretic consequences of a group that has a winning strategy for <script type="math/tex">I</script>, how about for <script type="math/tex">II</script>?</li>
</ul>Paul PlummerIn my post on the strong measure zero game, I mentioned a game , where player one, who I will call , presents player two, who I will call , a subset from a group , where the normal closure of the set is the full group, and , chooses one element from each set presented, trying to get a set whose normal closure is again the full group. The game lasts innings ( if you prefer that notation). Essentially the idea is that gives presentations of the trivial group, and is trying to end up with a presentation of the trivial group.The Strong Measure Zero Game2015-05-15T00:00:00+00:002015-05-15T00:00:00+00:00http://mathsalad.com/2015/05/15/the-strong-measure-zero-game<h2 id="background">Background</h2>
<p>Originally, my first post was going to be an introduction to selection principles, especially those in topology, and a few other related concepts. I decided I could not do justice to the subject in one post (especially with my minimal knowledge), and have opted to present some very cool results that will, hopefully, give a taste of the ideas that come up when studying the subject.</p>
<p><strong>Definition of the game <script type="math/tex">G_1( \cal A, \cal B)</script>.</strong> Let <script type="math/tex">\mathcal{A}</script> and <script type="math/tex">\mathcal{B}</script> be collections of sets, we define the game <script type="math/tex">G_1(\cal{A}, \cal{B})</script> to be a game with two players, that we will denote <script type="math/tex">I</script> and <script type="math/tex">II</script>, and the game will have <a href="https://en.wikipedia.org/wiki/Ordinal_number"><script type="math/tex">\omega</script> innings</a>, where each inning consists of a turn by <script type="math/tex">I</script> and then a turn by <script type="math/tex">II</script>. By <script type="math/tex">\omega</script> innings we mean there is an inning for each <script type="math/tex">0,1,2,3, \dots</script> (if you prefer <script type="math/tex">\omega = \mathbb N</script>). The innings follow this pattern: player <script type="math/tex">I</script> starts off, in inning <script type="math/tex">n</script>, by presenting a nonempty set <script type="math/tex">A_n</script> from <script type="math/tex">\cal A</script> which marks the end of player <script type="math/tex">I</script>’s turn, and then player <script type="math/tex">II</script> picks an <script type="math/tex">a_n \in A_n</script>. At the end of the <script type="math/tex">\omega</script> innings we are left with a play</p>
<script type="math/tex; mode=display">(A_0,a_0, \dots , A_i, a_i, \dots)</script>
<p>of what happened in that game. Player <script type="math/tex">II</script> wins if <script type="math/tex">\{ a_i \mid i \in \omega \} \in \cal B</script> and <script type="math/tex">I</script> wins otherwise.</p>
<p>A couple of examples to consider:</p>
<ul>
<li>Let <script type="math/tex">D</script> consist of all dense subsets of <script type="math/tex">\mathbb R</script>. Consider <script type="math/tex">G_1 (D,D)</script>.</li>
<li>Let <script type="math/tex">X</script> be a countable subset of some topological space <script type="math/tex">Y</script> and <script type="math/tex">{\cal{O}}_Y</script> is the set of all open covers of <script type="math/tex">Y</script>, and <script type="math/tex">{\cal {O}}_X</script> the set of covers for <script type="math/tex">X</script>. Consider <script type="math/tex">G_1( {\cal{O}}_Y, {\cal{O}}_X)</script>.</li>
<li><script type="math/tex">V</script> is a finite dimensional rational vector space, and <script type="math/tex">A</script> is the set of infinite vector subspaces of <script type="math/tex">V</script>, and <script type="math/tex">B</script> is the set of sets that contain infinite rational vector subspaces of <script type="math/tex">V</script>. Consider <script type="math/tex">G_1(A,B)</script></li>
<li>Let <script type="math/tex">U</script> be an <a href="https://en.wikipedia.org/wiki/Ultrafilter">ultrafilter</a> on some set. Consider <script type="math/tex">G_1(U,U)</script>.</li>
<li>Let <script type="math/tex">H</script> be a group and <script type="math/tex">{\mathcal N}_H</script> the set of subsets of <script type="math/tex">H</script> that normally generate the group. Consider <script type="math/tex">G_1({\mathcal N}_H, {\mathcal N}_H)</script> . There are also variation of considering sets that generate the group <script type="math/tex">H</script>, <script type="math/tex">{\mathcal K}_H</script> and looking at <script type="math/tex">G_1({\mathcal K}_H, {\mathcal N}_H)</script>, <script type="math/tex">G_1({\mathcal N}_H, {\mathcal K}_H)</script>, <script type="math/tex">G_1({\mathcal K}_H, {\mathcal K}_H)</script> (there is a very good chance I am going to write about this bullet point, and related ideas, in future posts, since it is basically the reason I started the blog).</li>
<li>etc. Think of some others!</li>
</ul>
<p>A strategy for a player is a function that takes what has been played and outputs the next move. So <script type="math/tex">s_1,s_2</script> will be strategies for players <script type="math/tex">I,II</script> respectively, and <script type="math/tex">s_1: \varnothing \mapsto A_0</script>, and <script type="math/tex">s_1: (A_0,a_0,...,A_i,a_i) \mapsto A_{i+1}</script>. We get a similar description for <script type="math/tex">s_2</script>. Basically they are instructions on a game tree that tell the player what to do next given where they are in the game tree. We call <script type="math/tex">s_i</script> a winning strategy provided the cooresponding player will <em>always</em> win following that strategy. It is interesting that in some games there is no winning strategy for either player, and in fact the two main theorems that will be presented will give such an example(assuming <a href="https://en.wikipedia.org/wiki/Strong_measure_zero_set">Borel’s conjecture</a> is false)</p>
<p>Do any of the above (bulleted)examples have winning strategies for one of the players? How about the games you came up with?</p>
<p><strong>Definition of strong measure zero.</strong> A set <script type="math/tex">X \subseteq \mathbb R</script> is strong measure zero if, for all sequences <script type="math/tex">( \epsilon_n \mid n \in \omega)</script> of positive real numbers, then there exists a sequence of intervals <script type="math/tex">I_n</script>, where the length of <script type="math/tex">I_n</script> is less than <script type="math/tex">\epsilon_n</script> for all <script type="math/tex">n</script>, and <script type="math/tex">X \subseteq \cup_{n \in \omega} I_n</script>.</p>
<p>Every strong measure zero set is <a href="https://en.wikipedia.org/wiki/Lebesgue_measure">Lebesgue measure zero</a>, although <a href="http://math.stackexchange.com/questions/903973/the-cantor-set-is-not-strong-measure-zero">not every measure zero set is strong measure zero</a>, the <a href="https://en.wikipedia.org/wiki/Cantor_set">Cantor set</a> provides such an example. It is also easy to see that every countable set is strong measure zero. In fact <a href="https://en.wikipedia.org/wiki/Strong_measure_zero_set">Borel’s conjecture</a> actually conjectures that every strong measure zero set is countable. That conjecture ended up being independent of the usual axioms, ZFC, so it is consistent for there to be uncountable strong measure zero sets, and consistent for there to be only countable strong measure zero sets. <a href="https://en.wikipedia.org/wiki/Luzin_space">Luzin sets</a> provide examples of strong measure zero sets that are not countable.</p>
<p>Let <script type="math/tex">J_{\epsilon}</script>, <script type="math/tex">\epsilon >0</script>, be the set of all intervals of length less than <script type="math/tex">\epsilon</script>, and say <script type="math/tex">{\mathcal J}:= \{J_\epsilon \mid \epsilon >0 \}</script>, and <script type="math/tex">{\mathcal O}_X</script> the set of open covers of some set <script type="math/tex">X \subseteq \mathbb R</script> (where the open sets come from <script type="math/tex">\mathbb R</script> and not <script type="math/tex">X</script> with the subspace topology).</p>
<p><strong>Definition of the strong measure zero game.</strong> Let <script type="math/tex">X \subseteq \mathbb R</script>. Then the strong measure zero game is the game <script type="math/tex">G_1( {\mathcal J}, {\mathcal O}_X)</script>.</p>
<p>There are a couple of other definitions I could have chosen, but I want to stress that the game has a very strong connection to strong measure zero, since player <script type="math/tex">I</script> can choose a sequence of <script type="math/tex">\epsilon_n</script> and play <script type="math/tex">J_{\epsilon_n}</script>, in inning <script type="math/tex">n</script>, and then <script type="math/tex">II</script> tries to cover <script type="math/tex">X</script> by collecting one interval from each <script type="math/tex">J_{\epsilon_n}</script>.</p>
<p>A very natural question to ask is when does player <script type="math/tex">I</script> have a winning strategy, and when does player <script type="math/tex">II</script> have a winning strategy in <script type="math/tex">G_1( {\mathcal J}, {\mathcal O}_X)</script>? This brings us to “the main event”:</p>
<p><strong>Theorem 1.</strong> Player <script type="math/tex">II</script> has a winning strategy in the strong measure zero game, if, and only if, <script type="math/tex">X</script> is countable.</p>
<p><strong>Theorem 2.</strong> Player <script type="math/tex">I</script> has a winning strategy in the strong measure zero game, if, and only if, <script type="math/tex">X</script> is not strong measure zero.</p>
<p>One of the things that makes these results interesting is that this means it is <em>consistent</em> for winning strategies to always exist, that is when all strong measure zero sets are countable, and it is consistent for there to be sets where there is no winning strategy, which is when there are uncountable strong measure zero sets(as an example the Luzin sets mentioned before).</p>
<h2 id="the-proofs-of-theorem-1-and-2">The Proofs of Theorem 1 and 2</h2>
<p>The proofs of Theorem 1 and 2, that will be presented, is Galvin’s with some modification and can be found in his paper <em>Indeterminacy of point-open games</em>. I found out about this proof in a class, taught by Marion Scheepers, a few years ago.</p>
<p>First, some notation, <script type="math/tex">^\omega \omega</script> will be the set of infinite sequences of order type <script type="math/tex">\omega</script>, <script type="math/tex">% <![CDATA[
^{< \omega} \omega %]]></script> will be the set of finite sequences indexed by finite ordinals. And if <script type="math/tex">x,y</script> are sequences, then <script type="math/tex">x \frown y</script> will be the concatenation of the sequences. As examples</p>
<script type="math/tex; mode=display">(1,2,3,...) \in {^\omega \omega},</script>
<script type="math/tex; mode=display">% <![CDATA[
(7,10,2) \in {^{<\omega}} \omega %]]></script>
<p>and</p>
<script type="math/tex; mode=display">(7,10,2) \frown (24,1)= (7,10,2,24,1).</script>
<p><strong>Lemma 3.</strong> Let <script type="math/tex">X</script> be a set and suppose that for each <script type="math/tex">x \in X</script> we have that a set <script type="math/tex">B_x</script>, which is a set of positive integer, finite length, sequences such that:</p>
<ol>
<li>For every infinite sequence <script type="math/tex">(n_1,...,n_i,...)</script> there is a <script type="math/tex">k \in \omega</script> such that <script type="math/tex">(n_1,...,n_k) \in B_x</script>.</li>
<li>
<p>For each finite sequence <script type="math/tex">(n_1,...,n_m)</script>,</p>
<script type="math/tex; mode=display">\vert \bigcap_{k \in \omega} \{ x \in X \mid (n_1,...,n_m,k) \in B_x \} \vert \leq \aleph_0,</script>
</li>
</ol>
<p>then we have that <script type="math/tex">\vert X \vert \leq \aleph_0</script> (<script type="math/tex">X</script> is countable).</p>
<p><em>proof of Lemma 3.</em> We will prove by contradiction, so suppose that <script type="math/tex">X</script> is an uncountable set with the above properties. This means that</p>
<script type="math/tex; mode=display">% <![CDATA[
X \setminus \bigcup_{ \alpha \in { ^{< \omega} \omega} } \left( \bigcap_{n \in \omega} \{ x \in X \mid \alpha \frown (n) \in B_x \} \right) \neq \varnothing %]]></script>
<p>since the part that is being removed from <script type="math/tex">X</script> is countable. Let <script type="math/tex">x</script> be an element from this nonempty set. By the definition of the above set, we know that <script type="math/tex">x</script> has the property, that for any sequence <script type="math/tex">% <![CDATA[
\alpha \in {^{<\omega} \omega} %]]></script>, that there is an <script type="math/tex">n</script> such that <script type="math/tex">\alpha \frown (n) \not\in B_x</script>. Starting with the empty sequence, <script type="math/tex">()</script>, we know there is a least <script type="math/tex">n_1</script> such that <script type="math/tex">(n_1) \not\in B_x</script>, and we can continue doing this inductively, constructing sequences <script type="math/tex">(n_1,...,n_k) \not\in B_x</script> of length <script type="math/tex">k</script>. This contradicts property 1. <script type="math/tex">\square</script></p>
<p><strong>Theorem 1.</strong> Player <script type="math/tex">II</script> has a winning strategy in the strong measure zero game, if, and only if, <script type="math/tex">X</script> is countable.</p>
<p><em>proof of Theorem 1.</em> It is easy to see that <script type="math/tex">II</script> has a winning strategy, when <script type="math/tex">X</script> is countable, since we can enumerate the set <script type="math/tex">X</script>, and in turn <script type="math/tex">n</script>, <script type="math/tex">II</script> covers the <script type="math/tex">n</script>-th element. For the other direction, we will need to define a different game. Let <script type="math/tex">(Y,d)</script> be a separable metric space, so it also has a countable basis. (Separable means contains a countable dense set subset.) The game has these rules:</p>
<ol>
<li>There is an inning for each <script type="math/tex">n \in \omega</script>.</li>
<li>In inning <script type="math/tex">n</script> player <script type="math/tex">I</script> chooses a set <script type="math/tex">\{S_1,S_2\}</script>, where <script type="math/tex">S_i \subseteq Y</script> is open and the distance between <script type="math/tex">S_1</script> and <script type="math/tex">S_2</script> is positive, where distance between the two sets is defined to be <script type="math/tex">\inf \{ d(x,y) \mid (x,y) \in S_1 \times S_2 \}=d(S_1,S_2)</script>. We will call such pairs playable</li>
<li>Player <script type="math/tex">II</script> chooses one of the sets, which we will call <script type="math/tex">T_n</script>.</li>
<li>Player <script type="math/tex">II</script> wins if <script type="math/tex">\bigcap_{n \in \omega} T_n = \varnothing</script>.</li>
</ol>
<p>Lets call this game <script type="math/tex">\Gamma(Y)</script>, or the disjoint open ball game on <script type="math/tex">Y</script>. We will now show that when <script type="math/tex">II</script> has a winning strategy for <script type="math/tex">\Gamma(Y)</script> then the set <script type="math/tex">Y</script> is countably infinite. Let <script type="math/tex">\sigma</script> be a winning strategy for <script type="math/tex">\Gamma(Y)</script>. Note that we may assume that <script type="math/tex">I</script> only plays from sets from a basis, and since we have a countably basis, we may assume that <script type="math/tex">I</script> plays from some countable basis. Let <script type="math/tex">W= \{ P_n \mid n \in \omega \}</script> where <script type="math/tex">(P_n)_{n \in \omega}</script> is an enumeration of all playable <script type="math/tex">\\{ S_1, S_2\\}</script>, where <script type="math/tex">S_i</script> are from our chosen countable basis. For <script type="math/tex">x \in Y</script> we define</p>
<script type="math/tex; mode=display">% <![CDATA[
\begin{align*}
B_x := \{ (n_1,...,n_k ) \in {^{<\omega} \omega} \mid &x \in \sigma(P_{n_1}),...., \\
&x \in \sigma(P_{n_1},T_{n_1},...,P_{n_{k-1}}), \\
&x \not\in \sigma(P_{n_1},....,P_{n_{k-1}},T_{n_{k-1}},P_{n_k}) \}
\end{align*} %]]></script>
<p>Given that <script type="math/tex">\sigma</script> is a winning strategy for <script type="math/tex">II</script>, we can see that for any <script type="math/tex">x \in B_x</script> and sequence <script type="math/tex">(n_1,...) \in {^\omega \omega}</script> that some segment <script type="math/tex">(n_1,...,n_k) \in B_x</script>, since otherwise <script type="math/tex">x \in \sigma(P_{n_1},...,P_{n_k})=T_{k-1}</script> for all <script type="math/tex">k>0</script>, so <script type="math/tex">x \in \bigcap_{i>0} T_i</script> contradicting that set being empty. Hence we have condition 1, from Lemma 3. We will now show condition 2 of Lemma 3. Suppose that we have distinct</p>
<script type="math/tex; mode=display">x_1,x_2,x_3 \in \sigma( P_{n_1},....,P_{n_k} )</script>
<p>for <script type="math/tex">1 \leq k \leq m</script>. We can choose playable pairs</p>
<script type="math/tex; mode=display">P_{n_{1,2}}=\{A_1,A_2\},P_{n_{1,3}}=\{B_1,B_3\},P_{n_{2,3}}= \{C_2,C_3\}</script>
<p>where the index <script type="math/tex">i</script> on the open sets says that it contains <script type="math/tex">x_i</script> but not the <script type="math/tex">x_j</script> where <script type="math/tex">j \neq i</script>, since we are in a metric space. And note that only one of the <script type="math/tex">x_i=y_1</script> can be in
<script type="math/tex">\sigma(P_{n_1},...,P_{n_m},P_{n_{k,l} })</script>, hence other two, <script type="math/tex">y_2,y_3</script>, are not contained in that set. This is true for each combination, so we have</p>
<script type="math/tex; mode=display">(n_1,...,n_m,n_{k,l}) \in \left(B_{y_2} \cap B_{y_3} \right)\setminus B_{y_1}.</script>
<p>So we get that <script type="math/tex">y_1 \not \in \{ x \mid (n_1,....,n_m,n_{k,l}) \in B_x</script>, and by shuffling things around, we get that each of the three points, <script type="math/tex">x_1,x_2,x_3</script>, have cooresponding sets that they are not in, hence they are not in the intersection of all those sets. This worked for an arbitrary choice of three points, so it must be</p>
<script type="math/tex; mode=display">\bigcap_{n \in \omega} \{x \in Y \mid (n_1,...,n_m,n) \in B_x \}= \varnothing.</script>
<p>Using Lemma 3, we get that the space is countable.</p>
<p>Let <script type="math/tex">\Sigma</script> be a winning strategy for <script type="math/tex">II</script> in the strong measure zero game on <script type="math/tex">X</script>. We are going to show that the strategy <script type="math/tex">\Sigma</script> will help us construct a winning strategy <script type="math/tex">\sigma</script> in the game <script type="math/tex">\Gamma(X)</script>, hence <script type="math/tex">X</script> will be countable.</p>
<p>Say that player <script type="math/tex">I</script> plays <script type="math/tex">P_1=\{S_1,S_2\}</script> in <script type="math/tex">\Gamma(X)</script>, and plays <script type="math/tex">J_{d(S_1,S_2)}=J_{\epsilon_1}</script> in the strong measure zero game. We will have player <script type="math/tex">II</script> play <script type="math/tex">A_1=\Sigma(J_{\epsilon_1})</script>, and then <script type="math/tex">\sigma(P_1)</script> will be an element disjoint from <script type="math/tex">A_1</script>, which we can pick since the length of <script type="math/tex">A_1</script> is less than <script type="math/tex">d(S_1,S_2)</script>. So we can define a strategy where
<script type="math/tex">\sigma( \{S_1^n,S_2^n\} )= S_1^n</script> if <script type="math/tex">\Sigma
\left(J_{d(S_1^n,S_2^N)} \right) \cap S_1^n = \varnothing</script> and <script type="math/tex">\sigma \left(\{S_1^n,S_2^n\} \right)= S_2^n</script> otherwise. We will now show that this is in fact a winning strategy. Note that</p>
<script type="math/tex; mode=display">\left(\bigcap_{n \in \omega} \sigma(P_1,...,P_n) \right) \cap \left( \bigcup_{n \in \omega} A_n \right) = \varnothing</script>
<p>but the union part covers the whole space so the intersection on the left must be empty. So player <script type="math/tex">II</script> has a winning strategy for <script type="math/tex">\Gamma(X)</script> when <script type="math/tex">II</script> has a winning strategy for the strong measure zero game. Thus <script type="math/tex">X</script> is countable.<script type="math/tex">\square</script></p>
<p><strong>Lemma 4 (Lebesgue covering lemma).</strong> If <script type="math/tex">X</script> is a compact metric space, then for any open cover of <script type="math/tex">X</script>, there exists a postive real number <script type="math/tex">\delta</script>, the <em>Lebesgue number</em>, such that: for any <script type="math/tex">F \subseteq X</script> of diameter less than <script type="math/tex">\delta</script> there is a <script type="math/tex">U</script> in the open cover such that <script type="math/tex">F \subseteq U</script>.</p>
<p><em>proof of Lemma 4.</em> This is a well known theorem, so no proof will be given. <script type="math/tex">\square</script></p>
<p><strong>Theorem 2.</strong> Player <script type="math/tex">I</script> has a winning strategy in the strong measure zero game, if, and only if, <script type="math/tex">X</script> is not strong measure zero.</p>
<p><em>proof of Theorem 2.</em> First we will prove (<script type="math/tex">\Leftarrow</script>), so suppose that <script type="math/tex">X</script> is not strong measure zero. Hence there is a sequence <script type="math/tex">(\epsilon_n)</script> for <script type="math/tex">n \in \omega</script> such that, such that <script type="math/tex">X</script> for all sequences of intervals <script type="math/tex">(A_n)</script>, for <script type="math/tex">{n \in \omega}</script>, length of <script type="math/tex">A_n</script> is less than or equal too <script type="math/tex">\epsilon_n</script>. So have player <script type="math/tex">I</script> plays <script type="math/tex">J_{\epsilon_n}</script> in inning <script type="math/tex">n</script>, so player <script type="math/tex">II</script> can not cover the space, since it will only pick intervals less than <script type="math/tex">\epsilon_n</script> at inning <script type="math/tex">n</script>.</p>
<p>We will now prove (<script type="math/tex">\Rightarrow</script>), we will do the contrapositive, so we will show that if <script type="math/tex">X</script> is strong measure zero then <script type="math/tex">I</script> does not have a winning strategy. Let <script type="math/tex">\sigma</script> be an arbitrary strategy for player <script type="math/tex">I</script>, and <script type="math/tex">X</script> is strong measure zero. We will show that there is a play, by player <script type="math/tex">II</script> that will beat this strategy. Here are some things that we are going to assume, to make the problem a bit easier:</p>
<ol>
<li>That we are playing on strong measure zero sets, that are subsets of compact sets, and in fact we will assume to be working with <script type="math/tex">[0,1]</script>. This is enough since we can write <script type="math/tex">\mathbb {R}</script> as a countable union of compact sets, for example <script type="math/tex">\bigcup_{i \in \mathbb Z} [i,i+1]</script>, and partition the innings, into countably infinite many infinite sets (think of sets of the form <script type="math/tex">\{p^n \mid n \in \mathbb{N} \}</script>, <script type="math/tex">p</script> prime ), so we play the strong measure zero game on the <script type="math/tex">j</script>th compact set when the inning we are in is in the <script type="math/tex">j</script>th partition.</li>
<li>We will assume that the <script type="math/tex">J_\epsilon</script>’s that are played by player <script type="math/tex">I</script> have <script type="math/tex">\epsilon</script>’s of the form <script type="math/tex">1/n</script>, since for some <script type="math/tex">\epsilon</script> we can just move to a smaller number of the form <script type="math/tex">1/n</script> without losing any generality.</li>
<li>Let <script type="math/tex">s=(J_{\epsilon_1},...,J_{\epsilon_n},K_n)</script> be some sequence of turns, that is in the domain of player <script type="math/tex">I</script>’s strategy <script type="math/tex">\sigma</script>, then we define <script type="math/tex">\hat\sigma(s)= \epsilon \iff \sigma(s) = J_\epsilon</script>.</li>
</ol>
<p>We will now construct a play in which <script type="math/tex">II</script> which beats player <script type="math/tex">I</script> strategy <script type="math/tex">\sigma</script>. Say that <script type="math/tex">\sigma( \varnothing)= J_{1/n_1}</script>. Lets say <script type="math/tex">U_1 \subset J_{1/n_1}</script> is a finte open cover of <script type="math/tex">[0,1]</script> (which exists since <script type="math/tex">[0,1]</script> is compact). Consider <script type="math/tex">\min \{ \hat \sigma(K) \mid K \in U_1 \}</script>, and we will choose <script type="math/tex">m_1 \in \mathbb{N}</script> large enough so that:</p>
<ol>
<li><script type="math/tex">% <![CDATA[
n_1<m_1 %]]></script>,</li>
<li><script type="math/tex">% <![CDATA[
1/m_1 < \min \{ \hat \sigma(K) \mid K \in U_1 \} %]]></script>,</li>
<li>and <script type="math/tex">1/m_1</script> is a Lebesgue number of the cover <script type="math/tex">U_1</script>.</li>
</ol>
<p>Basically we are going to go down, and play every possible outcome of the strategy <script type="math/tex">\sigma</script> possible. Let <script type="math/tex">U_2 \subset J_{1/m_1}</script>, be a finite open cover, so in particular this <script type="math/tex">U_2</script> works as a subcover for <script type="math/tex">\sigma(K)</script> for <script type="math/tex">K \in U_2</script> since <script type="math/tex">% <![CDATA[
1/m_1 < \hat \sigma (K) %]]></script>. Then we are going to choose an <script type="math/tex">m_2 \in \mathbb N</script> large enough so that:</p>
<ol>
<li><script type="math/tex">% <![CDATA[
m_1 < m_2 %]]></script> (this also means that <script type="math/tex">% <![CDATA[
\hat \sigma ( K ) < m_2 %]]></script> for <script type="math/tex">K \in U_1</script>, which are the corresponding “n_i” terms ),</li>
<li><script type="math/tex">% <![CDATA[
1/m_2< \min\{ \hat \sigma (K_1,K_2) \mid K_1 \in U_1, K_2 \in U_2 \} %]]></script>,</li>
<li>And <script type="math/tex">1/m_2</script> is a Lebesgue number of the cover <script type="math/tex">U_2</script>.</li>
</ol>
<p>Basically we continue in this fashion, and construct <script type="math/tex">m_t</script>, for arbitrary <script type="math/tex">t</script> (which if that statement was good enough for you, I recommend skipping this paragraph). Suppose that we have <script type="math/tex">m_1,...,m_{t-1}</script> constructed. Then we are going to construct <script type="math/tex">m_t</script> by letting <script type="math/tex">U_t \subset J_{1/m_{t-1}}</script> be a finite open cover. Then we choose <script type="math/tex">m_t \in \mathbb{N}</script> large enough such that:</p>
<ol>
<li><script type="math/tex">% <![CDATA[
m_{t-1} < m_t %]]></script>,</li>
<li><script type="math/tex">% <![CDATA[
1/m_t < \min \{ \hat \sigma (K_1,...,K_t) \mid K_1 \in U_1,..., K_t \in U_t \} %]]></script>,</li>
<li>And <script type="math/tex">1/m_t</script> is a Lebesgue number of the cover <script type="math/tex">U_t</script>.</li>
</ol>
<p>So know we have a sequence <script type="math/tex">(1/m_i)_{i \in \omega}</script>, where each <script type="math/tex">1/m_i</script> is a Lebesgue number of the coverings <script type="math/tex">U_i</script>. Since <script type="math/tex">X</script> is strong measure zero we have that there is a covering <script type="math/tex">\{L_i \mid i \in \omega\}</script> where each <script type="math/tex">L_i</script> has length less than <script type="math/tex">1/m_i</script>. Since <script type="math/tex">1/m_i</script> is a Lebesgue number of the cooresponding coverings, we can find an <script type="math/tex">M_i \in U_i</script> such that <script type="math/tex">L_i \subseteq M_i</script>. We can choose the <script type="math/tex">M_i</script> as our sets, and so we found a play where <script type="math/tex">II</script> wins, hence <script type="math/tex">\sigma</script> is not a winning strategy, and since <script type="math/tex">\sigma</script> was arbitrary we are done.
<script type="math/tex">\square</script></p>
<h2 id="resources">Resources</h2>
<p>You can find out more about these sorts of things from this, <em>non-exhaustive</em>, list:</p>
<ul>
<li>F. Galvin, Indeterminacy of point-open games, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 26 :5 (1978), 445–449 (this is sort of difficult to find, so included all the information, for libraries)</li>
<li>“The combinatorics of open covers (I): Ramsey theory” by Marion Scheepers,</li>
<li>“The combinatorics of open covers (II)” by Winfried Just, Arnold W. Miller, Marion Scheepers, and Paul J. Szeptycki,</li>
<li>…</li>
<li>“The combinatorics of open covers (<script type="math/tex">N</script>)” by … (well there are at least 11 of these…),</li>
<li>“Ramsey theory of open covers lecture 1” by Nadav Samet and Boaz Tsaban,</li>
<li>…</li>
<li>“Ramsey theory of open covers lecture 5” by Nadav Samet and Boaz Tsaban.</li>
<li>You can browse around <a href="http://diamond.boisestate.edu/~spm/index.html">here</a></li>
</ul>
<p>The “Ramsey theory of open covers” are more accessible and shorter than “The combinatorics of open covers…” and discuss some of the Ramsey theoretic aspects that are related to these games and to “selection principles” (which I did not discuss here, although are very related to games <script type="math/tex">G(\mathcal A,B)</script>).</p>Paul PlummerBackground